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2 years ago

How do you find the Area of a pentagon

Mathematics

2 answers:

solmaris [256]2 years ago

8 0

Answer:

I think base times hight

Morgarella [4.7K]2 years ago

4 0

Answer:

Area of a regular pentagon = pa/2, where p = the perimeter and a = the apothem. If you don't know the perimeter, calculate it from the side length: p = 5s, where s is the side length.

Step-by-step explanation:

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-5 + 4 = -1

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djyliett [7]

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2 years ago

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3 years ago

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tensa zangetsu [6.8K]

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2 years ago

The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of

kompoz [17]

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

$A=\frac{\sqrt{3} }{4} a^{2}$

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

$A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}$

$A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}$

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

$A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}$

$A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}$

Now, comparing areas:

$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

$\frac{A_{1}}{A_{2}} =$ 2.25

The area of ΔABC is 2.25x greater than the area of ΔDEF.

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2 years ago